Problem: $\int\dfrac{e^x}{1+e^{2x}}dx\,=$ Choose 1 answer: Choose 1 answer: (Choice A) A $\tan^{-1}(e^x)+C$ (Choice B) B $\ln{(1+e^{x})+C}$ (Choice C) C $\ln{(1+e^{2x})+C}$ (Choice D) D $\sin^{-1}(e^x)+C$
Explanation: Notice that we can rewrite the integral as $ \int \dfrac{1}{1+(e^x)^2}\cdot \,e^x\, dx\,$. If we let $ {u=e^x}$, then $du=e^x \, dx}$. Substituting gives us: $ \int \dfrac{1}{1+({e^x})^2}\,\cdot e^x\, dx}\,= \int \dfrac{1}{1+ u^2}\, du}\,$ We recognize this antiderivative. $\begin{aligned}\phantom{\int\dfrac{e^x}{1+e^{2x}}dx}&= \int\dfrac{1}{1+u^2}\,du~\\\\\\ &=~ \tan^{-1}u+C\end{aligned}$ We can now substitute back to find the antiderivative in terms of $x$. ∫ e x 1 + e 2 x d x = tan − 1 u + C = tan − 1 ( e x ) + C \begin{aligned}\phantom{\int\dfrac{e^x}{1+e^{2x}}dx~}&=~\tan^{-1}u+C\\\\\\\ &=~\tan^{-1}(e^x)+C\end{aligned} The answer: $\int\dfrac{e^x}{1+e^{2x}}dx\,=~\tan^{-1}(e^x)+C$